Problem: $\dfrac{dy}{dx}=3y$, and $y=2$ when $x=1$. Solve the equation. Choose 1 answer: Choose 1 answer: (Choice A) A $y=2e^{3x+3}$ (Choice B) B $y=e^{3x-6}$ (Choice C) C $y=2e^{3x-3}$ (Choice D) D $y=2e^{3x+1}$
The general solution of equations of the form $\dfrac{dy}{dx}=ky$ is $y=C\cdot e^{kx}$ for some constant $C$. This can be found using separation of variables. In our case, $k=3$, so $y=C\cdot e^{3x}$. Let's use the fact that $y=2$ when $x=1$ to find $C$ : $\begin{aligned} y&=C\cdot e^{3x} \\\\ 2&=C\cdot e^{3\cdot 1} \gray{\text{Plug }x=1\text{ and }y=2} \\\\ 2e^{-3}&=C \end{aligned}$ In conclusion, $y=2e^{3x-3}$.